Greetings,

You're lucky. I've never had to write any paper for my math class (those that's just my naivety--I'm still in high school).

Differentiation of the equation in (10) yields a polynomial of odd (n - 1) degree. From the fundamental theorem of algebra (or some corollary of it), we realize that a polynomial of odd degree must have at least one real root, and thus the original function must have an absolute minimum (as the coefficient to is positive) along . For theorem 11, imagine the left hand side instead being set to f(x). Now, f(x) doesn't necessarily have to have any real solutions, as it is even. Concerningcin this function, the polynomial is shifted downcunits. As f(x) "points up" (as , and as , ), and if it has no real solutions, then the x-axis lies below the absolute minimum of f(x). Theorem (11) says thatthere exists a numbermsuch thatf(x) is moved beyond the x-axis, giving it a real solution. (It has to have an even number of real solutions and they can be multiplicative.) Ifcis greater than or equal to this numberm, then it has been shifted down the appropriate number of units to attain real solutions, but ifcis less than this numberm, then we haven't real solutions.

edit: I sorta implied it with "as the coefficient to is positive", but that and the remark about f(x) "pointing up" are directly related to one another. The sign would determine with way the polynomial "points". Also, there exists no loss of generality in the assumption that f(x) = c doesn't already have real solutions (hence the absolute minimum lies on or below the x-axis), for we're looking for a numbermthat satisfies the inequalities.